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I've been self-studying Rotman's An Introduction to the Theory of Groups, and Corollary 10.44 is that "an indecomposable [abelian] group $G$ is either torsion or torsion-free". The proof given is as follows:

Assume that $0 < tG < G$. Now $tG$ is not divisible, lest it be a summand of $G$, so that Corollary 10.43 shows that $G$ has a (cyclic) summand, contradicting indecomposability.

The referenced Corollary 10.43 states: A torsion [abelian] group $G$ that is not divisible has a $p$-primary cyclic direct summand (for some prime $p$).

Now, in the last step of this proof, I agree that $tG$ has this direct summand (it is the torsion group in the hypothesis of Corollary 10.43), but I am not sure why this implies it is also a summand of the whole group $G$.

I am looking for either an explanation of why this is the case, a different proof of the theorem, or a disproof (since I see from this question that the corresponding statement is not true for general modules over integral domains).

pancini
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I Eat Groups
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  • Not sure if this answers your question, but if $G$ is finitely generated then $tG$ is always a summand of $G$, so the result follows from $10.43$. – pancini Dec 23 '22 at 00:51
  • In that (finitely generated) case then we wouldn't even need to use the corollary, since that already contradicts indecomposability. I am looking for in the general case, though. – I Eat Groups Dec 23 '22 at 01:06

1 Answers1

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This is also using Corollary 10.41, which says a pure subgroup of bounded order is a direct summand. In particular, a cyclic direct summand of $tG$ will be pure in $G$, and thus also a direct summand of $G$.

Here is an alternate argument using a bit of homological machinery. First, note that if $A$ is a torsion-free abelian group, then $\operatorname{Ext}(A,B)$ is divisible for any $B$. This is because for any nonzero $n\in\mathbb{Z}$, multiplication by $n$ is an injection $A\to A$ which then induces a surjection $n:\operatorname{Ext}(A,B)\to \operatorname{Ext}(A,B)$ (here we crucially use the fact that $\mathbb{Z}$ is a PID so $\operatorname{Ext}^2(A/nA,B)=0$). In particular, then, if $B$ is torsion of bounded order (say, $n$), then the multiplication by $n$ map $\operatorname{Ext}(A,B)\to \operatorname{Ext}(A,B)$ is both surjective and equal to $0$, so $\operatorname{Ext}(A,B)=0$.

Now in your situation, suppose $C$ is a cyclic summand of $tG$. By the paragraph above, $\operatorname{Ext}(G/tG,C)=0$. Now consider the exact sequence $$\operatorname{Ext}(G/tG,C)\to\operatorname{Ext}(G/C,C)\to \operatorname{Ext}(tG/C,C).$$ The class of $G$ in $\operatorname{Ext}(G/C,C)$ has image $0$ in $\operatorname{Ext}(tG/C,C)$, since $C$ is a direct summand of $tG$. Since $\operatorname{Ext}(G/tG,C)=0$, this means the class of $G$ in $\operatorname{Ext}(G/C,C)$ is equal to $0$, i.e. $C$ is a direct summand of $G$.

Eric Wofsey
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